A Level Mathematics 9709 Paper 32, May/June 2024

Correct shape, roughly symmetrical. Both sections should be solid straight lines. 2a marked on each axis.

Solve linear equation or inequality to obtain critical value $x=\frac{5}{3}a$ or exact equivalent.

Obtain $x<\frac{5}{3}a$ or exact equivalent

State or imply the form $A+\frac{B}{2x+3}+\frac{C}{x-4}$

Use a correct method for finding a constant

Obtain one of $A=3$, $B=-2$ and $C=4$

Obtain a second value

Obtain a third value

Use logarithms to obtain a correct expression without powers e.g. $(2y-1)\ln a=(x-y)\ln b$

Separate terms and factorise to obtain $y(2\ln a+\ln b)=x\ln b+\ln a$

Clear explanation of linear form. From correct work only.

Use $a=b^3$ and log laws to simplify their equation

Obtain $y=\frac{1}{7}x+\frac{3}{7}$

Substitute $y=1$ and obtain ${\mathrm{e}}^x=a$, where $a>0$

Obtain ${\mathrm{e}}^x=2$ only

State or imply $\frac{\mathrm{d}}{\mathrm{d}x}(y{\mathrm{e}}^{2x})=2y{\mathrm{e}}^{2x}+{\mathrm{e}}^{2x}\frac{\mathrm{d}y}{\mathrm{d}x}$

State or imply $\frac{\mathrm{d}}{\mathrm{d}x}(y^2{\mathrm{e}}^x)=y^2{\mathrm{e}}^x+2y{\mathrm{e}}^x\frac{\mathrm{d}y}{\mathrm{d}x}$

Differentiate RHS of given equation to obtain zero (could be implied by subsequent work), substitute for $x$ and $y$ and obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\dots$

Obtain $-\frac{5}{4}$ or $-1.25$

Calculate the value of a relevant expression or values of a pair of expressions at $x=0.7$ and $x=0.8$

Complete the argument correctly with correct calculated values. (can be using the equation in the rubric or the equation in (b) or equivalent)

State $2x=\ln (5+\cos 3x)$ and take exponential of both sides to obtain ${\mathrm{e}}^{2x}=5+\cos 3x$

Use the iterative process correctly at least once

Obtain final answer 0.740

Show sufficient iterations to at least 5dp to justify 0.740 to 3dp, or show that there is a sign change in the interval $(0.7395,0.7405)$

Use correct product rule

Obtain correct derivative

Equate derivative to zero and solve for $x$

Obtain $x=\frac{1}{a},y=\frac{1}{a\mathrm{e}}$

Use integration by parts to obtain $px{\mathrm{e}}^{-ax}+q\int {\mathrm{e}}^{-ax}\mathrm{d}x$

Obtain $-\frac{1}{a}x{\mathrm{e}}^{-ax}+\frac{1}{a}\int {\mathrm{e}}^{-ax}\mathrm{d}x$

Complete integration to obtain $-\frac{1}{a}x{\mathrm{e}}^{-ax}-\frac{1}{a^2}{\mathrm{e}}^{-ax}$

Correct use of limits 0 and $\frac{2}{a}$ in an expression of the form $rx{\mathrm{e}}^{-ax}+s{\mathrm{e}}^{-ax}$

Obtain $\frac{1}{a^2}(1-3{\mathrm{e}}^{-2})$

Factorise LHS using difference of 2 squares

Simplify

Obtain ${\cos }^4\theta -{\sin }^4\theta \equiv \cos 2\theta$ from correct working

Use part (a) and correct double angle formula to obtain expression involving $\int {\sin }^{2}2\theta \mathrm{d}\theta$ or $\int {\cos }^{2}2\theta \mathrm{d}\theta$

$\int \cos 2\theta \mathrm{d}\theta =\frac{1}{2}\sin 2\theta$

Use of correct double angle formula on second part of the integral to obtain a form that can be integrated directly

Obtain $\frac{1}{2}\theta -\frac{1}{8}\sin 4\theta$

Correct use of limits $\pm \frac{\pi }{8}$ in an expression of the form $p\theta +q\sin 2\theta +r\sin 4\theta$ and evaluate the trig

Obtain $\frac{1}{2}\sqrt{2}+\frac{1}{8}\pi -\frac{1}{4}$

Correct direction vector seen or implied ($\overrightarrow{BC}=3\mathbf{i}+3\mathbf{j}-3\mathbf{k}$)

Use a correct method to form a vector equation

Obtain $\mathbf{r}=5\mathbf{i}+2\mathbf{j}+\lambda (\mathbf{i}+\mathbf{j}-\mathbf{k})$

Use components to form two relevant equations in 2 unknowns

Solve 2 relevant equations in 2 unknowns for $\lambda$ or $\mu$

Obtain $\lambda =2$ or $\mu =3$

Obtain $(7,4,-2)$

State $AB=\sqrt{7^2+1^2+4^2}(=\sqrt{66})$

State $\overrightarrow{BD}$ in component form

$AB=BD\Rightarrow (3r-7{)}^2+(r-1{)}^2+(-2r+4{)}^2=66$ $(14r^2-60r=0)$

$\Rightarrow r=\frac{30}{7}$

$\overrightarrow{OD}=\frac{76}{7}\mathbf{i}+\frac{37}{7}\mathbf{j}-\frac{32}{7}\mathbf{k}$

State $z\omega =(-3+3\sqrt{3})+(3+3\sqrt{3})\mathrm{i}$

Obtain $\mathrm{\mid }z\mathrm{\mid }=\sqrt{2}$

Obtain $\arg z=-\frac{\pi }{4}$ final answer

Obtain $\mathrm{\mid }\omega \mathrm{\mid }=6$

Obtain $\arg \omega =\frac{2\pi }{3}$ final answer

Show $\mathrm{\mid }OA\mathrm{\mid }=\mathrm{\mid }AB\mathrm{\mid }=6$, hence isosceles

$\mathrm{\angle }AOB=\arg \omega -\arg z\omega =-\arg z=\frac{\pi }{4}$ hence third angle is a right angle

$\arg z\omega =\arg z+\arg \omega (=\frac{2\pi }{3}-\frac{\pi }{4}=\frac{5\pi }{12})$

$\arg z\omega ={\tan }^{-1}\frac{3+3\sqrt{3}}{-3+3\sqrt{3}}$

$\Rightarrow \tan (\frac{5}{12}\pi )=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Use of correct chain rule (and correct quotient rule) and ${\cos }^{-3}\theta$

$\frac{\mathrm{d}y}{\mathrm{d}\theta }=-3\times -\sin \theta (\cos \theta {)}^{-4}=3\sin \theta {\sec }^4\theta$ Must be expressed in the given form

Separate variables: $\int \frac{\sin \theta }{{\cos }^4\theta }\mathrm{d}\theta =\int \frac{(x+3)}{(x^2+9)}\mathrm{d}x$

Obtain $p{\sec }^3\theta (+A)$

Use $\int \frac{3x+9}{x^2+9}\mathrm{d}x=\int (\frac{3x}{x^2+9}+\frac{9}{x^2+9})\mathrm{d}x$ and obtain $q\ln (x^2+9)$ or $r{\tan }^{-1}\frac{x}{3}(+C)$.

Obtain $q\ln (x^2+9)$ and $r{\tan }^{-1}\frac{x}{3}(+C)$

Obtain ${\sec }^3\theta =\frac{3}{2}\ln (x^2+9)+3{\tan }^{-1}\frac{x}{3}(+C)$ or equivalent

Use $\theta =\frac{1}{3}\pi ,x=3$ in an equation including $p{\sec }^3\theta$, $q\ln (x^2+9)$ and $r{\tan }^{-1}\frac{x}{3}$ to evaluate the constant of integration

Obtain constant $=8-\frac{3}{2}\ln 18-\frac{3}{4}\pi$

Obtain $\cos \theta =0.601$